3.105 \(\int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx\)

Optimal. Leaf size=149 \[ \frac {2 i a^2 \text {Chi}\left (x f+\frac {c f}{d}\right ) \sinh \left (e-\frac {c f}{d}\right )}{d}-\frac {a^2 \text {Chi}\left (2 x f+\frac {2 c f}{d}\right ) \cosh \left (2 e-\frac {2 c f}{d}\right )}{2 d}-\frac {a^2 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (2 x f+\frac {2 c f}{d}\right )}{2 d}+\frac {2 i a^2 \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{d}+\frac {3 a^2 \log (c+d x)}{2 d} \]

[Out]

-1/2*a^2*Chi(2*c*f/d+2*f*x)*cosh(-2*e+2*c*f/d)/d+3/2*a^2*ln(d*x+c)/d+2*I*a^2*cosh(-e+c*f/d)*Shi(c*f/d+f*x)/d+1
/2*a^2*Shi(2*c*f/d+2*f*x)*sinh(-2*e+2*c*f/d)/d-2*I*a^2*Chi(c*f/d+f*x)*sinh(-e+c*f/d)/d

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Rubi [A]  time = 0.35, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3318, 3312, 3303, 3298, 3301} \[ \frac {2 i a^2 \text {Chi}\left (x f+\frac {c f}{d}\right ) \sinh \left (e-\frac {c f}{d}\right )}{d}-\frac {a^2 \text {Chi}\left (2 x f+\frac {2 c f}{d}\right ) \cosh \left (2 e-\frac {2 c f}{d}\right )}{2 d}-\frac {a^2 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (2 x f+\frac {2 c f}{d}\right )}{2 d}+\frac {2 i a^2 \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{d}+\frac {3 a^2 \log (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Sinh[e + f*x])^2/(c + d*x),x]

[Out]

-(a^2*Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(2*d) + (3*a^2*Log[c + d*x])/(2*d) + ((2*I)*a^2*C
oshIntegral[(c*f)/d + f*x]*Sinh[e - (c*f)/d])/d + ((2*I)*a^2*Cosh[e - (c*f)/d]*SinhIntegral[(c*f)/d + f*x])/d
- (a^2*Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*d)

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

\begin {align*} \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx &=\left (4 a^2\right ) \int \frac {\sin ^4\left (\frac {1}{2} \left (i e+\frac {\pi }{2}\right )+\frac {i f x}{2}\right )}{c+d x} \, dx\\ &=\left (4 a^2\right ) \int \left (\frac {3}{8 (c+d x)}-\frac {\cosh (2 e+2 f x)}{8 (c+d x)}+\frac {i \sinh (e+f x)}{2 (c+d x)}\right ) \, dx\\ &=\frac {3 a^2 \log (c+d x)}{2 d}+\left (2 i a^2\right ) \int \frac {\sinh (e+f x)}{c+d x} \, dx-\frac {1}{2} a^2 \int \frac {\cosh (2 e+2 f x)}{c+d x} \, dx\\ &=\frac {3 a^2 \log (c+d x)}{2 d}-\frac {1}{2} \left (a^2 \cosh \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\cosh \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx+\left (2 i a^2 \cosh \left (e-\frac {c f}{d}\right )\right ) \int \frac {\sinh \left (\frac {c f}{d}+f x\right )}{c+d x} \, dx-\frac {1}{2} \left (a^2 \sinh \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\sinh \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx+\left (2 i a^2 \sinh \left (e-\frac {c f}{d}\right )\right ) \int \frac {\cosh \left (\frac {c f}{d}+f x\right )}{c+d x} \, dx\\ &=-\frac {a^2 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{2 d}+\frac {3 a^2 \log (c+d x)}{2 d}+\frac {2 i a^2 \text {Chi}\left (\frac {c f}{d}+f x\right ) \sinh \left (e-\frac {c f}{d}\right )}{d}+\frac {2 i a^2 \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (\frac {c f}{d}+f x\right )}{d}-\frac {a^2 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 117, normalized size = 0.79 \[ -\frac {a^2 \left (-4 i \text {Chi}\left (f \left (\frac {c}{d}+x\right )\right ) \sinh \left (e-\frac {c f}{d}\right )+\text {Chi}\left (\frac {2 f (c+d x)}{d}\right ) \cosh \left (2 e-\frac {2 c f}{d}\right )+\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )-4 i \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (f \left (\frac {c}{d}+x\right )\right )-3 \log (c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Sinh[e + f*x])^2/(c + d*x),x]

[Out]

-1/2*(a^2*(Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*f*(c + d*x))/d] - 3*Log[c + d*x] - (4*I)*CoshIntegral[f*(c/d
+ x)]*Sinh[e - (c*f)/d] - (4*I)*Cosh[e - (c*f)/d]*SinhIntegral[f*(c/d + x)] + Sinh[2*e - (2*c*f)/d]*SinhIntegr
al[(2*f*(c + d*x))/d]))/d

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fricas [A]  time = 0.50, size = 149, normalized size = 1.00 \[ -\frac {a^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac {2 \, {\left (d e - c f\right )}}{d}\right )} - 4 i \, a^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (\frac {d e - c f}{d}\right )} + 4 i \, a^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-\frac {d e - c f}{d}\right )} + a^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right )} - 6 \, a^{2} \log \left (\frac {d x + c}{d}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))^2/(d*x+c),x, algorithm="fricas")

[Out]

-1/4*(a^2*Ei(2*(d*f*x + c*f)/d)*e^(2*(d*e - c*f)/d) - 4*I*a^2*Ei((d*f*x + c*f)/d)*e^((d*e - c*f)/d) + 4*I*a^2*
Ei(-(d*f*x + c*f)/d)*e^(-(d*e - c*f)/d) + a^2*Ei(-2*(d*f*x + c*f)/d)*e^(-2*(d*e - c*f)/d) - 6*a^2*log((d*x + c
)/d))/d

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giac [A]  time = 0.16, size = 139, normalized size = 0.93 \[ -\frac {a^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac {2 \, c f}{d} - 2 \, e\right )} + 4 i \, a^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (\frac {c f}{d} - e\right )} - 4 i \, a^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (-\frac {c f}{d} + e\right )} + a^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (-\frac {2 \, c f}{d} + 2 \, e\right )} - 6 \, a^{2} \log \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))^2/(d*x+c),x, algorithm="giac")

[Out]

-1/4*(a^2*Ei(-2*(d*f*x + c*f)/d)*e^(2*c*f/d - 2*e) + 4*I*a^2*Ei(-(d*f*x + c*f)/d)*e^(c*f/d - e) - 4*I*a^2*Ei((
d*f*x + c*f)/d)*e^(-c*f/d + e) + a^2*Ei(2*(d*f*x + c*f)/d)*e^(-2*c*f/d + 2*e) - 6*a^2*log(d*x + c))/d

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maple [A]  time = 0.22, size = 193, normalized size = 1.30 \[ -\frac {i a^{2} {\mathrm e}^{-\frac {c f -d e}{d}} \Ei \left (1, -f x -e -\frac {c f -d e}{d}\right )}{d}+\frac {3 a^{2} \ln \left (d x +c \right )}{2 d}+\frac {a^{2} {\mathrm e}^{\frac {2 c f -2 d e}{d}} \Ei \left (1, 2 f x +2 e +\frac {2 c f -2 d e}{d}\right )}{4 d}+\frac {a^{2} {\mathrm e}^{-\frac {2 \left (c f -d e \right )}{d}} \Ei \left (1, -2 f x -2 e -\frac {2 \left (c f -d e \right )}{d}\right )}{4 d}+\frac {i a^{2} {\mathrm e}^{\frac {c f -d e}{d}} \Ei \left (1, f x +e +\frac {c f -d e}{d}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*sinh(f*x+e))^2/(d*x+c),x)

[Out]

-I*a^2/d*exp(-(c*f-d*e)/d)*Ei(1,-f*x-e-(c*f-d*e)/d)+3/2*a^2*ln(d*x+c)/d+1/4*a^2/d*exp(2*(c*f-d*e)/d)*Ei(1,2*f*
x+2*e+2*(c*f-d*e)/d)+1/4*a^2/d*exp(-2*(c*f-d*e)/d)*Ei(1,-2*f*x-2*e-2*(c*f-d*e)/d)+I*a^2/d*exp((c*f-d*e)/d)*Ei(
1,f*x+e+(c*f-d*e)/d)

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maxima [A]  time = 0.48, size = 150, normalized size = 1.01 \[ \frac {1}{4} \, a^{2} {\left (\frac {e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} E_{1}\left (-\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {2 \, \log \left (d x + c\right )}{d}\right )} + i \, a^{2} {\left (\frac {e^{\left (-e + \frac {c f}{d}\right )} E_{1}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} - \frac {e^{\left (e - \frac {c f}{d}\right )} E_{1}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} + \frac {a^{2} \log \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))^2/(d*x+c),x, algorithm="maxima")

[Out]

1/4*a^2*(e^(-2*e + 2*c*f/d)*exp_integral_e(1, 2*(d*x + c)*f/d)/d + e^(2*e - 2*c*f/d)*exp_integral_e(1, -2*(d*x
+ c)*f/d)/d + 2*log(d*x + c)/d) + I*a^2*(e^(-e + c*f/d)*exp_integral_e(1, (d*x + c)*f/d)/d - e^(e - c*f/d)*ex
p_integral_e(1, -(d*x + c)*f/d)/d) + a^2*log(d*x + c)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sinh(e + f*x)*1i)^2/(c + d*x),x)

[Out]

int((a + a*sinh(e + f*x)*1i)^2/(c + d*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \frac {\sinh ^{2}{\left (e + f x \right )}}{c + d x}\, dx + \int \left (- \frac {2 i \sinh {\left (e + f x \right )}}{c + d x}\right )\, dx + \int \left (- \frac {1}{c + d x}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))**2/(d*x+c),x)

[Out]

-a**2*(Integral(sinh(e + f*x)**2/(c + d*x), x) + Integral(-2*I*sinh(e + f*x)/(c + d*x), x) + Integral(-1/(c +
d*x), x))

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